CHAPTER II

SINUSOIDAL STEADY STATE ANALYSIS

“Alternating Current Circuits (A.C.)”

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Objectives

• To review basic facts about sinusoidal signals.

• To introduce Phasors and convert the time domain

sinusoidal waveform into Phasors.

• To develop the phasor relationships for the basic

circuit elements.

• To solve electric circuits in phasor domain.

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Sinusoids

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A periodic function is one that satisfies v(t) = v(t + nT), for all t and for all integers n.

2T

HzT

f1

f 2

• Only two sinusoidal values with the same frequency can be compared by

their amplitude and phase difference.

• If phase difference is zero, they are in phase; if phase difference is not zero,

they are out of phase.

Phasor

• Transform a sinusoid to and from the time domain to the phasor domain:

(time domain) (phasor domain)

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)cos()( tVtv m mVV

• Amplitude and phase difference are two principal concerns in the study

of voltage and current sinusoids.

• Phasor will be defined from the cosine function in all our proceeding

study.

• If a voltage or current expression is in the form of a sine, it will be

changed to a cosine by subtracting from the phase.

Phasor

The differences between v(t) and V:

• v(t) is instantaneous or time-domain representation

V is the frequency or phasor-domain representation.

• v(t) is time dependent, V is not.

• v(t) is always real with no complex term, V is generally

complex.

Note: Phasor analysis applies only when frequency is

constant; when it is applied to two or more sinusoid

signals only if they have the same frequency.

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Sinusoid-Phasor Transformation

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The Phasor

• A phasor is a complex number that

represents the amplitude and phase

of a sinusoid.

• It can be represented in one of the

following three forms:

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rz

jrez

)sin(cos jrjyxz a. Rectangular

b. Polar

c. Exponential22 yxr

x

y1tanwhere

Example:

Transform the following sinusoids to phasors:

i = 6 Cos(50t – 40o) A

v = –4 Sin(30t + 50o) V

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Solution:

a. I A

b. Since –sin(A) = cos(A+90o);

v(t) = 4cos (30t+50o+90o) = 4cos(30t+140o) V

Transform to phasor => V V

406

1404

Circuit Elements in Frequency Domain

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VV

i

I

0 t

RV

icos( )mi I t

cos( )

cos( )

m

m

v Ri RI t

V t

rms rmsV RI

rms m

rms m

V VR

I I

Then：

1. Resistance

IV

O0

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V, i have the same

frequency，V leads i

by 90o

i

Vπ 2π

0 ωt

2. Inductance

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Relationship between voltage and current

sinmi I t

0cos sin( 90 )m m

div L LI t V t

dt

rms rmsV LI m mV LI

Reactance XL

sinmi I t 0sin( 90 )mv V t

Sinusoidal response of Inductance

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Effective Value

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i

v C

sinmv V t

0

0

sin( 90 )

sin( 90 )

m

m

dvi C C U t

dt

I t

Current leads voltage by 90°

2m m mI CV fCV Vi

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3. Capacitors

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Relationship between voltage and current

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Summary of voltage-current relationship

Element Time domain Frequency

domain

R

L

C

Riv RIV

dt

diLv LIjV

dt

dvCi

Cj

IV

Example

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Answer: i(t) = 30 cos(100t + 60o) mA

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It is given that the frequency of sinusoidal source is 50Hz, the peak value is 10V,

capacitor is 25μF, determine the peak value of current. If the frequency is 5000Hz,

then what is the peak value of current now？

i

V C

when f =50Hz

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1 1127.4

2 2 3.14 50 (25 10 )CX

fC

mAAX

V

C

m 78078.04.127

10Im

When f =5000Hz

6

1 11.274

2 2 3.14 5000 (25 10 )CX

fC

AX

VI

C

mm 8.7

274.1

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The higher frequency is under fixed peak value of voltage, the bigger the peak

value of current flowing through capacitor.

Solution

Example

Phasor Domain Sources

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• Convert time domain elements and sources into phasors

Time Domain Phasor Domain

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The Impedance

Kirchhoff’s Laws in the Frequency Domain

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• Both KVL and KCL are hold in the phasor domain or more

commonly called frequency domain.

• Moreover, the variables to be handled are phasors, which

are complex numbers.

• Series and parallel combinations are the same as in D.C.

circuits analysis.

• All the mathematical operations involved are now in

complex domain.

Example

Refer to Figure below, determine v(t) and i(t).

Answers: i(t) = 1.118sin(10t – 26.56o) A; v(t) = 2.236sin(10t + 63.43o) V

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Impedance Combinations

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Determine the input impedance of the circuit in figure below

at ω =10 rad/s.

Answer: Zin = 32.38 – j73.76

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